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(b) w = mg
CHAPTER 3
m _
2)
(
3

= (1.0 × 10 kg) 9.8 3.1. List the known and unknown quantities:
s
m _
3
= (1.0 × 10 ) (9.8) kg × F = 200 N

s 2
d = 3 m
kg·m
3 _ W= ?

= 9.8 × 10
s 2

These are the quantities found in the equation for work,
3
= 9.8 × 10 N W = Fd, which is already solved for work (W). Th us,
2.29. w = mg W = Fd
m _
2 )
(
kg·m
(
2)
= (70.0 kg) 9.8 = 200 _ (3 m)

s s
m _

= 70.0 × 9.8 kg × = (200)(3) N·m
s 2
= 600 J
kg·m
_
= 686 3.2. List the known and unknown quantities:
s 2
= 690 N F = 440 N
d = 5.0 m
mv
_ 2
2.30. F =
r w = 880 N
m _ 2 W = ?
(0.20 kg)( 3.0 )

s
__ These are the quantities found in the equation for work,

=
1.5 m W = Fd, which is already solved for work (W). As you can
(
2)
2
m _

(0.20 kg) 9.0 see in the equation, the force exerted and the distance the
s
__ box was moved are the quantities used in determining the
=

1.5 m work accomplished. The weight of the box is a diff erent
0.20 × 9.0 _ _
_ kg·m 2 1 variable, and one that is not used in this equation. Th us,

= × W = Fd

1.5 s 2 m
kg·m
2 )
kg·m·m ( _
_

= 1.2 = 440 (5.0 m)
2
s ·m s
= 1.2 N = 2,200 N·m
= 2,200 J
2.31. (a) Newton’s laws of motion consider the resistance to a change
3.3. Note that 10.0 kg is a mass quantity, and not a weight quantity.
of motion, or mass, and not weight. The astronaut’s mass is

Weight is found from w = mg, a form of Newton’s second law
kg·m

_ of motion. Thus, the force that must be exerted to lift the

1,960.0 2
2
s
w _ __ backpack is its weight, or (10.0 kg) × (9.8 m/s ), which is 98 N.

w = mg ∴ m = =

m _

g 9.8 Therefore, a force of 98 N was exerted on the backpack through
s 2 a distance of 1.5 m, and
1,960.0 kg·m s 2 W = Fd
_ _ _

= × = 200 kg kg·m

2 )
(
9.8 s 2 m = 98 _ (1.5 m)

(b) From Newton’s second law of motion, you can see that the s
= 147 N·m
100 N rocket gives the 200 kg astronaut an acceleration of
= 150 J
kg·m
_ 3.4. Weight is defined as the force of gravity acting on an object,

100
2
s
F _ _ and the greater the force of gravity, the harder it is to lift the

F = ma ∴ a = =
m 200 kg object. The force is proportional to the mass of the object, as

the equation w = mg tells you. Thus, the force you exert when

100 kg·m
_ 1 _ 2

= lift ing is F = w = mg, so the work you do on an object you lift

× = 0.5 m/s
200 s 2 kg
must be W = mgh.
2
(c) An acceleration of 0.5 m/s for 2.0 s will result in a fi nal You know the mass of the box, and you know the work
velocity of accomplished. You also know the value of the acceleration due to
gravity, g, so the list of known and unknown quantities is
_
v f – v i

a = ∴ v f = at + v i m = 102 kg

t 2
2
= (0.5 m/s )(2.0 s) + 0 m/s g = 9.8 m/s
W = 5,000 J
= 1 m/s
h = ?
650 APPENDIX E Solutions for Group A Parallel Exercises E-8